answer:Answer: A=-11, A=0.Solution. Let f(x)=a x+b, g(x)=a x+c, where a neq 0. The tangency of the graphs y=(f(x))^{2} and y=11 g(x) is equivalent to the equation (f(x))^{2}=11 g(x) having exactly one solution. We get (a x+b)^{2}=11(a x+c), a^{2} x^{2}+a(2 b-11) x+b^{2}-11 c=0. The discriminant of this equation is a^{2}(2 b-11)^{2}-4 a^{2}left(b^{2}-11 cright)=11 a^{2}(11-4 b+4 c), from which 4 b-4 c=11.Similarly, the tangency of the graphs y=(g(x))^{2} and y=A f(x) means that the equation (g(x))^{2}=A f(x) has a unique solution. This equation is equivalent to the following: (a x+c)^{2}=A(a x+b), a^{2} x^{2}+a(2 c-A) x+c^{2}-A b=0. The discriminant is D=a^{2}(2 c-A)^{2}-4 a^{2}left(c^{2}-A bright)=a^{2} A(A- 4 c+4 b). It becomes zero when A=0 and A=4 c-4 b=-11.

answer:Answer: Yes.Solution. For example, frac{1}{200}<frac{2}{300}. There are many other examples.Criteria. Any correct example: 7 points.Answer without an example or incorrect answer: 0 points.

answer:Among the statements on the paper, any two contradict each other, so at most one statement can be true. The 100th statement cannot be true, because otherwise it would contradict itself. If the 100th statement is false, then there is a true statement among the statements, so according to our previous remark, there is exactly one true statement, and the other 99 are false. It is precisely this fact that the 99th statement records, so among the statements on the paper, the 99th statement is true, and the rest are false.

answer:7. It is evident that arranging of A in increasing order does not diminish m. Thus we can assume that A is nondecreasing. Assume w.l.o.g. that a_{1}=1, and let b_{i} be the number of elements of A that are equal to i left(1 leq i leq n=a_{2001}right). Then we have b_{1}+b_{2}+cdots+b_{n}=2001 and m=b_{1} b_{2} b_{3}+b_{2} b_{3} b_{4}+cdots+b_{n-2} b_{n-1} b_{n} Now if b_{i}, b_{j}(i<j) are two largest b's, we deduce from (1) and the AMGM inequality that m leq b_{i} b_{j}left(b_{1}+cdots+b_{i-1}+b_{i+1}+cdots+b_{j-1}+b_{j+1}+b_{n}right) leq left(frac{2001}{3}right)^{3}=667^{3} quadleft(b_{1} b_{2} b_{3} leq b_{1} b_{i} b_{j}right., etc.). The value 667^{3} is attained for b_{1}=b_{2}=b_{3}=667 (i.e., a_{1}=cdots=a_{667}=1, a_{668}=cdots=a_{1334}=2, a_{1335}=cdots=a_{2001}=3 ). Hence the maximum of m is 667^{3}.

answer:According to the concept of mapping:- A bijection is also referred to as a one-to-one correspondence, which is correct.- Since a bijection is injective, different elements in set A have different images in set B.- Since a bijection is surjective, every element in set B has a preimage in set A.- The set of images is the same as set B because a bijection is both injective and surjective.Therefore, the correct option is boxed{D}.A mapping that is both injective and surjective is called a bijection, which is also known as a one-to-one correspondence. This allows us to confirm options 1, 2, 3, and 4.This question examines the concept of a bijection and is considered a fundamental problem.

answer:SolutionESince angle S T Q=154^{circ} and angles on a straight line add to 180^{circ}, angle P T Q=26^{circ}. Hence, since P Q and Q T are equal, triangle P T Q is isosceles and so angle T P Q=26^{circ}. Therefore, since angles in a triangle add to 180^{circ}, angle T Q P=128^{circ}. Finally, since Q T and R S are parallel and corresponding angles on parallel lines are equal, angle S R Q=128^{circ}.